Q38E - University Physics with Modern Physics

Question

 Two loudspeakers, $\mathit{A}$ and  $\mathbf{B}$, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is $\mathbf{172}\mathbf{Hz}$. You are $\mathbf{8}\mathbf{.}\mathbf{00}\text{\hspace{0.17em}}\mathbf{m}$ from $\mathit{A}$ . What is the closest you can be to $\mathit{B}$ and be at a point of destructive interference?

Step-by-Step Solution

Answer

The closet distance from  $B$  is $1\mathrm{m}$.

Step 1: Given Data

Frequency of sound emitted:  $f=172\text{\hspace{0.17em}Hz}$

The distance of observer from A: $l=8\text{\hspace{0.17em}m}$

Step 2: Condition of destructive interference

Condition of destructive interference is $\mathbf{\Delta l}\mathbf{=}\frac{n}{2}\mathbf{\lambda }$ ,  $\mathbf{(}\mathbf{n}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{5}\mathbf{,}\mathbf{7}\mathbf{,}\mathbf{9}\mathbf{,}...\mathbf{)}$where $\mathbf{\Delta l}$ is the path difference and $\mathit{\lambda }$ is the wavelength of the sound emitted from the speakers.

Step 3: Use the condition

Given that $f=172\text{\hspace{0.17em}Hz}$ and we know that the speed of sound in air is $v=344\text{\hspace{0.17em}m/s}$ .

So,

 $\begin{array}{c}\lambda =\frac{v}{f}\\ =\frac{344\text{\hspace{0.17em}m}/\text{s}}{172\text{\hspace{0.17em}Hz}}\\ =2\text{\hspace{0.17em}m}\end{array}$

The two speakers are $8\mathrm{m}$ apart. Let the distance from B is $x$. So, we can write $\Delta l=8-x$.

Now,

 $\begin{array}{c}8-x=\frac{n}{2}\lambda \\ 8-x=\frac{n}{2}\times 2\\ 8-x=n\\ x=8-n\end{array}$

From the given values of  $n$, for $n=7$, we get the closest distance.

Hence, the closest distance from $B$  is $1m$.